\(\sqrt{\left(\sqrt{8}-4\right)^2}+\sqrt{8}=\left|\sqrt{8}-4\right|+\sqrt{8}=4-\sqrt{8}+\sqrt{8}=4\)
\(\left(5\right)\sqrt{x+3-4\sqrt{x-1}}\sqrt{x+8+6\sqrt{x-1}}=5\)
\(\left(6\right)2x^2+3x+\sqrt{2x^2+3x+9}=33\)
\(\left(7\right)\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+30}=8\)
\(\left(8\right)x+y+z+8=2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}\)
\(\left(6\right)\dfrac{3\sqrt{x}}{5\sqrt{x}-1}\le-3\)
\(\left(7\right)\dfrac{8\sqrt{x}+8}{6\sqrt{x}+9}>\dfrac{8}{3}\)
\(\left(8\right)\dfrac{\sqrt{x}-2}{2\sqrt{x}-3}< -4\)
\(\left(9\right)\dfrac{4\sqrt{x}+6}{5\sqrt{x}+7}\le-\dfrac{2}{3}\)
\(\left(10\right)\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}>-6\)
\(P\left(x\right)=\sqrt[3]{\sqrt{x+8}\left(x^4+8x^3+12x\right)+6x^3+48x^2+8}\)
đặt \(A=\sqrt{x+8}\left(x^4+8x^3+12x\right)+6x^3+48x^2+8\)
\(=\sqrt{x+8}\left(x^4+8x^3\right)+6x^2\left(x+8\right)+12x\sqrt{x+8}+8\)
\(=\sqrt{\left(x+8\right)^3}x^3+3\sqrt{\left(x+8\right)^2}x^22+3\sqrt{\left(x+8\right)}x4+8\)
\(=\left(x\sqrt{x+8}+2\right)^3\)
\(\Rightarrow P\left(x\right)=x\sqrt{x+8}+2\)
\(\sqrt{\left(2x+3\right)^2}=5\)
\(\sqrt{9.\left(x-2\right)^2}=18\)
\(\sqrt{9x-18}-\sqrt{4x-8}+3\sqrt{x-2}=40\)
\(\sqrt{4.\left(x-3\right)^2}=8\)
\(\sqrt{4x^2+12x+9}=5\)
\(\sqrt{5x-6}-3=0\)
Giải Phương Trình
\(\sqrt{\left(2x+3\right)^2}=5\)
\(\sqrt{9\left(x-2\right)^2}=18\)
\(\sqrt{9x-18}-\sqrt{4x-8}+3\sqrt{x-2}=40\)
\(\sqrt{4.\left(x-3\right)^2}=8\)
\(\sqrt{5x-6}-3=0\)
\(\left\{{}\begin{matrix}8\sqrt{xy-2y}-8y+4=\left(x-y\right)^2\\2\sqrt{2y-y^2}\left(\sqrt{8-2x}-2\sqrt{2y}+1\right)=4y+5\sqrt{2-y}-10\sqrt{x-2}\end{matrix}\right.\)
Chứng minh đẳng thức
\(\left(4-\sqrt{7}\right)^2=23-8\sqrt{7}\)
\(\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\)
\(\dfrac{\sqrt{4-2\sqrt{3}}}{1+\sqrt{2}}:\dfrac{\sqrt{2}-1}{\sqrt{3}+1}=2\)
\(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}=-1,5\)
Chứng minh đẳng thức
\(\left(4-\sqrt{7}\right)^2=23-8\sqrt{7}\)
\(\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\)
\(\dfrac{\sqrt{4-2\sqrt{3}}}{1+\sqrt{2}}:\dfrac{\sqrt{2}-1}{\sqrt{3}-1}=2\)
\(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}=-1,5\)
\(\sqrt{\dfrac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\dfrac{4}{\left(2+\sqrt{5}\right)^2}}=8\)
\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(\left(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\right)^2\)
\(4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right).\left(4-\sqrt{10+2\sqrt{5}}\right)}\)
\(8+2\sqrt{16-10-2\sqrt{5}}\)
\(8+2\sqrt{6-2\sqrt{5}}\)
\(8+2\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(8+2\left(\sqrt{5}-1\right)\)
\(8+2\sqrt{5}-2\)
\(6+2\sqrt{5}\)
\(\left(\sqrt{5+1}\right)^2\)
\(\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(|\sqrt{5}+1|\)
\(\sqrt{5}+1\)
\(\text{thư ngu như chó}\)
